Chapter 12.4, Problem 29AYU

Show that the formula

$2+4+6+···+2n=n^2+n+2$

obeys Condition II of the Principle of Mathematical Induction. That is, show that if the formula is true for some $k$, it is also true for $k+1$. Then show that the formula is false for $n=1$ (or for any other choice of $n$).

To determine

To prove: The given statement $2 + 4 + 6 + \cdots + 2n = n^2 + n + 2$ obeys the Condition II of Principle of Mathematical Induction and the statement is false for $n = 1$.

Answer to Problem 29AYU

The given statement is not true for $n = 1$

Explanation of Solution

Given:

Statements says $2 + 4 + 6 + \cdots + 2n = n^2 + n + 2$ obeys the Condition II of Principle of Mathematical Induction and the statement is false for $n = 1$.

Formula used:

The Principle of Mathematical Induction

Suppose that the following two conditions are satisfied with regard to a statement about natural numbers:

CONDITION I: The statement is true for the natural number 1.

CONDITION II: If the statement is true for some natural number $k$, it is also true for the next natural number $k + 1$. Then the statement is true for all natural numbers.

Proof:

Consider the statement

$2 + 4 + 6 + \cdots + 2n = n^2 + n + 2$    -----(1)

Step 1: Show that statement (1) is true for Condition II.

Assume that the statement is true for some natural number k.

That is $2 + 4 + 6 + \cdots + 2k = k^2 + k + 2$    -----(2)

Step 2: Prove that the statement is true for the next natural number $k + 1$.

That is, to prove that $2 + 4 + 6 + \cdots + 2\left(k + 1\right)={\left(k + 1\right)}^2 + \left(k + 1\right) + 2$

Consider $2 + 4 + 6 + \cdots + 2\left(k + 1\right)$

$= 2 + 4 + 6 + \cdots + 2k + 2\left(k + 1\right)$

$= k^2 + k + 2 + 2\left(k + 1\right)$

$= k^2 + 3k + 4$

RHS ${\left(k + 1\right)}^2 + \left(k + 1\right) + 2 = k^2 + 2k + 1 + k + 1 + 2 = k^2 + 3k + 4$

Step 3: Prove that the statement is not true for $n = 1$

$2\left(1\right) \ne {\left(1\right)}^2 + 1 + 2$

Therefore the given statement is not true for $n = 1$